4 reds in a row: the odds — both of them

Two different questions hide in this query. A specific window: the next 4 spins all red = (18/37)⁴ = 5.601% (1 in 18). Your actual session: the chance of seeing a run of 4+ reds at least once is 96.2% in 100 spins, 99.9% in 200, and it’s practically certain — computed exactly with a Markov chain (method), a figure most sites never touch.⚙ computed

Will you see it? Session odds

Session length	P(streak ≥ 4 at least once)	Expected count of such runs
50 spins	79.64%	1.38
100 spins	96.17%	2.82
200 spins	99.86%	5.69
500 spins	100.00%	14.32
1000 spins	100.00%	28.70

Four-runs arrive about twice per 200-spin session on average — frequent enough that “the table is streaky tonight” is usually just Tuesday.

Why this streak matters: it’s a Martingale killer

Betting black through 4 reds on a $5 Martingale means your next bet is $80 and climbing. Our million-session Martingale run shows exactly how often these walls arrive.

The fallacy checkpoint

After 4 reds, the next spin is still red with probability 18/37 = 48.65%. The wheel has no memory — streaks are what independence looks like, not evidence against it. The 1913 Monte Carlo table famously ran 26 blacks while the room bet ever-heavier on red; the episode named the gambler’s fallacy — its place in the game’s full timeline here.

Other streak lengths

3 in a row 5 in a row 6 in a row 7 in a row 8 in a row 9 in a row odds hub →

Frequently asked

What are the odds of 4 reds in a row in roulette?

For a fixed window: (18/37)^4 ≈ 5.601% on a European wheel, about 1 in 18. On American wheels it’s rarer: (18/38)^4 ≈ 5.034%.

Will I ever see 4 reds in a row?

More often than intuition says: at least once in 100% of 200-spin sessions by exact Markov computation. Long streaks are a normal feature of independent spins.

After 4 reds, is black due?

No — each spin is independent, so black is still 48.65% (EU). Betting bigger on “due” colors is the gambler’s fallacy, and it’s exactly how streaks eat Martingale bankrolls.